A 20 liter rigid tank contains 19 kg of R134 at 24 C What is

A 20 liter rigid tank contains 1.9 kg of R-134 at 24 C. What is the specific volume, pressure and state of the R-134 (calculate the quality if applicable) How much R-134 must be added/removed so that the contents are a saturated liquid at the same temperature?

Solution

(a) For R-134 at 24 degree C,

Saturation pressure = 645.8 kPa

Specific volume

vf = 0.0008261 m^3/kg

vg = 0.0319 m^3/kg

--> Specific volume corresponding to 20 liter and 1.9 kg of gas is, v = (20*10^-3)/1.9 = 0.01053 m^3/kg

--> Pressure will be same as saturation pressure = 645.8 kPa

--> The state is wet vapour. Let the dryness fraction be x.

then, vf + x*(vg-vf) = 0.01053, so x = 0.312 or 31.2%

(b) As the volume doesn\'t change so specific volume is v = vf = 0.0008261 m^3/kg

So mass of R-134 required = V/v = (20*10^-3)/0.0008261 = 24.21 Kg

So the mass that needs to be added = 24.21 - 1.9 = 22.31 kg.